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Artist 52 [7]
3 years ago
7

How many total electrons are shared between two nitrogen atoms when they bond to form a diatomic nitrogen molecule?

Chemistry
1 answer:
GuDViN [60]3 years ago
8 0

Explanation:

Diatomic Nitrogen molecule means the two atoms of nitrogen N_{2}.

  • A single nitrogen contains 5 electrons in its valence shell which has 3 single electrons and one pair of electron. When the bond is being made the three single electrons makes the covalent bond with the 3 electrons of other atom which results in 3 pairs . The total electrons shared between the atoms are 3 pairs  (6 electrons) . while when it completes the octate the N_{2} is formed.
  • Nitrogen is colorless, odorless, tasteless and the inert diatomic gas.Nitrogen combines of near about 78% of Earth's atmosphere.It is non-flammable which does not support combustion , it occurs in all living organisms. Nitrogen is used in most of the Biological processes  and also used in the formation of fertilizers, in the form of ammonia and ammonia based compounds.It is also used in Chemical Industries for making nylon , explosives, dyes and etc. The breathing of Nitrogen is deadly, as it displaces the oxygen.
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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =
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<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

[O_2]_{eq}=0.587M

K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

Hence, the equilibrium concentration of water is 0.597 M

8 0
3 years ago
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